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Full text of "Simplified design of concrete floor systems : with design tables."

SIMPLIFIED DES1QN 

or 

CONCRETE FLOOR 

SYSTEMS 



With Design Table* 




PORTLAND CEMENT ASSOCIATION 



SIMPLIFIED DE5IQN 



OF 

CONCRETE 



FLOOR 



SYSTEMS 



With Design Tables 



Published by 

PORTLAND CEMENT ASSOCIATION 

33 WEST GRAND AVENUE • CHICAGO. ILLINOIS 






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41 






TAB LES 



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r 






Table No. Subject Page 



1 Bending Moments in Slabs, Beams and Girders for Uni- 

form Loads .,.,.,.. 



3 



14 



2 Summary of Quantities and Panel Loads. .. . .39 

3 Values of Design Constants for Various Combinations of 

Steel and Concrete Stresses ....... 39 

4 Rectangular Beams with Compression Reinforcement 

Design Values of p and K ....... 40 



5 Sectional Area, in Sq. In., of Various Numbers of Bars . 40 

6 Area of Steel, in Sq. In. per Linear Foot, of Slab for Bars 

Spaced At Various Intervals * . - . 41 

7 One- Way Solid Concrete Slabs . . . 50 

8 Concrete or Clay Tile and Concrete Joists . . . 52 



9 20-in. Metal Pans and Concrete Joists (Straight Pans) 

10 20-in. Metal Pans and Concrete Joists (Tapered Pans) . 56 

11 30-in. Metal Pans and Concrete Joists (Straight Pans) . 58 

12 30-in. Metal Pans and Concrete Joists (Tapered Pans) . 60 

13 Concrete T-Bcams (Simple Spans) * . . . 62 

14 Concrete T-Beams (Two Spans) . ..... 66 

15 Concrete T-Beams (End Spans) . , . . . . 67 

16 Concrete T-Beams (Interior Spans) ...... 68 

17 Two- Way Solid Concrete Slabs (Square Panels) 70 

18 Two-Way Solid Concrete Slabs (Ratio of Spans 1.2 to 1.0) 72 

19 Two- Way Solid Concrete Slabs (Ratio of Spans 1.4 to 1.0) . 74 

20 Two-Way Solid Concrete Slabs (Ratio of Spans 1.6 to 1.0) 76 

21 Two-Way Solid Concrete Slabs (Ratio of Spans 1.8 to 1.0) . 78 

22 Two-Way Solid Concrete Slabs (Ratio of Spans 2.0 to 1.0) . 80 

23 Concrete Stair Slabs ........ 83 

24 Recommended Live Loads ...... 85 



25 Weights of Building Materials ...... 86 



To those who design 

ECONOMIC FLOORS 



T 



HE availability and economy of materials 
used in reinforced concrete make this form of construction 
ideal for floors for all uses- The very nature of concrete per- 
mits a flexibility in the size and arrangement of floor mem- 
bers. It materially lowers insurance rates. 

Concrete floors are rigid and have great load carrying capac- 
ity when correctly designed. With the proper finish these floors 
arc easily kept clean, and wear is negligible. Their long, useful 
life with minimum maintenance cost produces an ideal eco- 
nomic combination. 

This booklet enumerates and discusses various concrete 
floor systems, shows their respective advantages and relative 
economies, and illustrates the method of design by typical 
examples. Different floor constructions suitable for a set of 
specific requirements are analyzed lo illustrate the procedure 
in determining the most economical type. 

Safe load tables of numerous types of slabs and beams are 
iven to assist the designer by simplifying the calculations 



involved. These tables are based on requirements in the build- 
ing code of the American Concrete Institute, as this code has 
come into quite general use. Differences between this code 
and local building codes, if any, must necessarily be taken 
into account by the designer. 



Portland Cement Association 



4 



Simplified Design 



of 



Concrete Floor Systems 



SECTION I— CONCRETE FLOOR SYSTEMS 
AND THEIR CHARACTERISTICS 



THE choice of a type of concrete floor for any specific use involves the 
careful consideration of points characteristic of each system. Intelli- 
gent comparisons may then be made and selection of a suitable, economical 
floor becomes a relatively simple matter. The elimination of types not 
inhereDtly adapted to the proposed use reduces the number for which 
comparative cost studies should be made. 

The most important points to be considered are : 

/. Use of Floor 

1. Loading. 

a. Light or heavy. 

b. Uniform or concentrated. 

2. Span, 

a. Long or short. 

b. Regularity of support arrangement. 

3. Freedom from obstruction. 

a. Columns. 

b. Beams or girders. 

c. Incorporation of pipes, conduits or vent shafts. 

4. Rigidity and resistance to: 

a. Machinery shocks. 

b. Outside vibrations. 

c. Wind stresses or earthquakes. 

5. Adaptability to alteration. 

a. Main members. 

b. Secondary members. 



6 



Simplified Design of 



6. Conformity to code. 

a. City or other local codes. 

b. Fire or other insurance codes. 

7. Insulation. 

a. Sound. 

b. Heat or cold. 

8. Acoustical properties. 

II. Cost of Floor 

1. First cost. 

2. Maintenance cost. 

3. Indirect cost. (Affecting cost of remainder of building.) 

a. Ability to "tie in" to structure. 

b. Difference in weight and story height as affecting size and cost of 
supporting columns and foundations. 

4. Royalty costs of patented types. 

5. Salvage of formwork. 

///. Appearance of Floor 

1. Floor finish. 

a. Integral concrete. 

b. Any finish applied later. 

2. Ceiling finish. 

a. Exposed and untreated. 

b. Plaster. 

c. Paint. 

IV* Local Building Conditions 

1. Material. 

a. Supply, including freight rate. 

b. Handling on the job. 

c. Storage space. 

2. Labor. 

a. Type available. 

b. Possibility of importation. 

c. Labor restrictions. 

3. Special conditions of bidding by contractors. 

4. Climatic conditions. 



Concrete Floor Systems 



7 



V, Time Requirements 

1. Speed of construction as affecting revenue of the building. 

2. Delays caused by purchase of special materials not available locally. 

VI* Supervision 

1. Normal. 

2. Extra rigid, due to unusual or new system involved. 

3. Ease of repairing the result of slight errors or omissions, without 
reducing factor of safety. 

Several of the more commonly used systems are enumerated in the 
pages to follow, with descriptions, general characteristics and advantages 
for special purposes or unusual conditions. To facilitate discussion and 
without intention of fixing definite limits, live loading for floors, as referred 
to in this review, will be divided into three groups and defined as light 
(from 40 or 50 to 100 p. si.*), intermediate (100 to 175 p.s.f.) and heavy 
(any live load in excess of 175 p.s.f.). 

One-Way Solid Slab 

Concrete slabs of uniform depth, with no filler material, and with the 
main reinforcemeDt in one direction only, arc called one-way solid slabs. 
The term "flat slab" is often erroneously applied to such slabs. 

♦p.s.f. is an abbreviation of "pounds per square foot." 




One-way solid slab construction produces rigid, economical floors suitable for 

industrial plants and buildings of other occupancy. 



8 



Simplified Design of 



Constructed in this manner such slabs constituted the earliest applica- 
tion of concrete to a floor system. They have many good points which 
make them suitable and economical when used for supporting intermediate 
and heavy loads on comparatively short spans, say, from 6 to 12 ft. long. 
For light loads these spans may be increased to 14 or 15 ft. 

Construction progresses rapidly, as all of the materials used are easily 
procured, and only ordinary labor and normal supervision are necessary. 

Solid slabs produce a stiff and rigid floor capable of withstanding vibra- 
tion and shock from machinery or earthquakes, and are adapted to use 
where heavy concentrated loads must be supported. 

They provide an excellent base for concrete floor finishes, are easily kept 
clean and are allowed a low insurance rate because of their high fire 
resistive qualities. 

Limitations to the use of one-way solid slabs for long spans are due to 
their comparatively large dead weight. Also, the introduction of more 
beams than used in other systems may not meet the architectural require- 
ments for the floor layout below. The slab weight may be considerably 
reduced by the use of lightweight aggregate. This reduction may effect a 
pronounced saving in the indirect cost of the structure (i.e., in reducing 
the sizes of columns and footings). Lightweight aggregate also improves 
heat insulation. 

If good tight forms are constructed of pressed wood fiber boards or 
similar material, plaster may often be omitted, as an attractive ceiling 
surface results from merely painting or staining the exposed concrete. 

Ordinary electric conduits and pipes of about the same size can generally 
be run in the slab, but mechanical equipment of Larger size may require 
space provided for bj ? either a suspended ceiling or an extra fill between 
the slab and the finished floor. 



One-Way Ribbed Slabs with Block Filler 

Hollow filler blocks of lightweight concrete or clay tile, when laid in 
rows in the bottom of concrete slabs, constitute one form of ribbed slab. 
The dead load is considerably reduced as compared with a solid slab of 
equal load carrying capacity although the total depth of slab is increased. 
The width of the concrete joists, separating the rows of filler and encasing 
the reinforcement, may be made any desired width to meet the strength 
requirements. It is customary to include a solid concrete top of two inches 
or more in depth over the blocks. This serves the double purpose of pro- 
viding a space for concealing small pipes and conduits and together with 
the joists gives a T-section, thereby adding considerable strength to them. 
Introduction of the filler blocks improves sound and heat insulation of slab. 

While plaster may be applied directly to the slab, it has been noted that 
at limes a slight discoloration takes place when clay tile filler is used. This 
is due to the use of materials of different densities and absorption qualities 
next to one another and may be eliminated by placing tile soffit piecefl in 
I he bottoms of the joists to form an all-tile ceiling. 



Concrete Floor Systems 



9 




Ribbed slabs constructed with removable metal pans make a lightweight floor 

of pleasing appearance. 




Ribbed slabs of this type are well adapted to floors of medium span, with 
light and intermediate loadings. They cannot carry heavy concentrated 
loads as well as the solid slabs, and become relatively less economical 
when the spans are increased. 

To insure straight joists, care should be taken that the filler blocks are 
maintained accurately in line. Owing to the porous nature of the blocks, 
should be thoroughly sprinkled to prevent absorption of the water 
in the concrete, particularly in warm weather. 

One-Way Ribbed Slabs with Metal Pans 

One of the lightest types of concrete floors results from the use of metal 
pan fillers between concrete joists. Numerous kinds of such pans are 
available. Some are fabricated of light gauge material and are intended to 
be left in place. They may be procured with metal lath already attached 
at the bottom, upon which plaster may be applied directly. Other pans 
are made of heavier metal and are removed when the forms are stripped, 
to be used again or returned to the owner, if leased. Both styles are fur- 
nished with either straight or tapered closed ends. Tapered pans are highly 
desirable on long spans, as with their use the width of the joists is increased 
at points where such width is needed, namely, near the supports where the 
shearing stresses are highest. The usual width of metal pans is 20 inches, 
although 10, 15 and 30-in. pans are generally available. 



10 



Simplified Design of 




Kihhed slabs often have spreader joists at third points of long spans. They arc 

eeonomical for light loads. 

Metal pan ribbed floors reach their highest economy on long spans with 
light loads. While the weight is very light, the depth of the floor produce 
great rigidity. Form work is economical, merely requiring a board under 
• ach rib or joist, with the space under the pans left open. 

These floors are not so well suited to support concentrated loads, as 
the lopping between joists is comparatively thin (2J^ to 3 inches). Care 
must be taken to reinforce the topping across construction joints to pre- 
vent these joints from opening. 

Two-Way Solid Slab 

Solid concrete slabs reinforced in two directions, and carrying load both 
w.iys to marginal beams, offer a system that is quite economical for 
certain uses and has other decided advantages. Modern building codes 
prea nt improved design methods which make this type of floor economical 
lid // suited to support intermediate and heavy loads on spans up to about 
90 ft. Best results are obtained when panels are nearly square, although 
panels having a ratio of long to short sides of 2 to 1 may be designed. 

Marginal beams framing in two directions to columns, while presenting 
possible architectural difficulty as to floor layout below, produce effective 
horizontal bracing. Wind and earthquake forces as well as vibrating 
machinery loads are properly resisted, and the slab is well suited to carry 
heavy concentrated loads through so-called "plate action." 



Concrete Floor Systems 



11 



Other advantages of the two-way slab are its high fire-resistive qualities, 
simplicity of formwork and ease of erection. If lightweight aggregate is 
used, the dead load may be reduced about 30%. This reduction will be 
reflected in lowering the indirect costs throughout the building structure. 

Two- Way Ribbed Slabs with Block Filler 

The introduction of hollow filler blocks of lightweight concrete or clay 
tile, with concrete joists running in both directions, results in a type of 
floor having nearly all the advantages of both the one-way ribbed slab and the 
two-way solid slab. Dead load is considerably reduced, but the total floor 




One-way and two-way ribbed slabs with block fillers produce good surfaces for 

plastering. Note the soffit blocks beneath the concrete ribs. 

thickness is increased. Care should be taken with this type, if clay tile 
filler is used, that the concrete is of such consistency as not to flow into 

and fill up the cells in the tile. 

Other remarks made for one-way ribbed slabs with block filler will in 

general apply to the two-way ribbed slabs. 

Girderless or "Flat Slab" Floors 

Solid concrete slabs reinforced in two or more directions, but without 
marginal beams of any kind, are called "flat slabs." They have all the 
economy, high efficiency and rigidity of the two-way slabs previously men- 



12 



Simplified Design of 




Maximum story height without obstruction is obtained with flat slab floor 

construction. 




Painted flat slab ceilings arc* suitable for department stores arid industrial 

buildings and are economical for the heaviest loads. 



Concrete Floor Systems 



13 



tioned, but in even higher degree. The absence of beams offers the additional 
advantages of better lighting, ventilation and arrangement of mechanical 
equipment if used in factory types of buildings. It is customary to employ 
supporting columns with flared heads or capitals. Also in some cases, it 
will be found that economy results from the use of panels of additional 
thickness at each column, called "drop panels." 

Flat slabs make rigid, efficient floors for such heavy and concentrated 
loads as are encountered in warehouses, loft buildings and industrial build- 
ings with heavy or vibrating machinery. Panels of medium span and nearly 
square are preferable. The slab must be made continuous over two or 
more spans. 

Being girderless, the flat slab requires the simplest formwork only and 
is given a high fire-resistive rating due to the absence of projecting corners 
or edges. Story heights may be considerably reduced on account of the 
greater available clear ceiling height. 



Summary 

While the previously described floors by no means exhaust the available 
types from which to choose, it will be found that in the majority of cases, 
at least one of those mentioned will provide an adequate, suitable and 
economical system; one that will give many years of service at a negligible 
maintenance cost. 



SECTION II— ELEMENTS OF DESIGN 



Prior to the actual design of a floor, the designer should be thoroughly 
familiar with the structural provisions of the building code he is to use. 
The safe load tables and other data given in this section and in Section 
III are based on provisions of the Tentative Building Regulations for 
Reinforced Concrete, American Concrete Institute,* referred to hereafter 
as the A. C. I. code, and concrete having an ultimate compressive strength 
of 2,000 p.s.i.** at an age of 28 days. This concrete has an allowable unit 
stress of 800 p.s.i. on the extreme fiber, with an increase to 900 p.s.i. on the 
extreme fiber adjacent to supports of continuous or fixed beams and slabs. 
The unit stress of the reinforcement is taken at 20,000 p.s.i. in tension, and 
n = 15. The shearing stress in the concrete is limited to 40 p.s.i. without 
web reinforcement and bond stress to 100 p.s.i. for deformed bars. Allow- 

♦Report of American Concrete Institute Committee 501, Standard Building Code, adopted by the Insti- 
tute at its 32nd annual convention, Feb. 1936, as a tentative standard. 

**p.a.i. is a generally accepted abbreviation of "pounds per square inch." 



14 



Simplified Destgn of 



able tensile stress in the web reinforcement is taken at 20,000* p.s.i. In all 
cases, the amount of protective covering of the reinforcing steel is shown 
in the tables. Bending moments and shears for continuous beams and 

slabs with uniform load are given in Table 1. 



TABLE No. 1— BENDING MOMENTS IN SLABS, BEAMS AND 

GIRDERS FOR UNIFORM LOADS 

Coefficients of wfi 



Number and Length of Spans 


End Spans 


Intermediate Spans 


+M 
at center 


-A£ 

at 2nd support I 


at center 


at support 


Spans 
equal 
to or less 
than 
10 ft. 




1/10 


1/10 


a ■ i ■ 


* * • * 


More than two spans 


1/10 


1/12 


1/12 


1/12 


Spans 
greater 
than 
1 ft. 


Two spans 


1/10 


1/8 


* * 1 » 


ft * • ■ 


More than two spans 


1/10 


1/10 


1/12 


1/12 

1 



¥ 1 

Note Shear in end members at the 2nd support is 1 .20 — and the shear at ail other supports is — 



For bending moments and shears in continuous girders with concen- 
trated loads, the coefficients given in Figure 1 are used. 

The coefficients in Table 1 and Figure 1 may be used for spans in which 
the longer of two adjacent spans does not exceed the shorter by more 
than 20 per cent, and in which the intensity of the live load is not more 
than three times the intensity of the dead load. In these cases, I equals 
the average of two adjacent clear span lengths for negative moment and 
the applicable clear span for positive moment. 

Negative reinforcement to be provided at the outer end of all members 
built integrally with their supports shall be not less than 0.005 b'd for 
T-beams and 0.005 bd for rectangular beams and slabs. 

A. C. I. Code recommendations with regard to "slabs supported on four 
sides" are the basis of values in the tables of two-way solid slabs. They 
take into account the condition of continuity of the slab, the ratio of the 
long side to the short side in rectangular panels, and the so-called "plate 
action" of the slab. 



•Sec footnote-** on page- 22 



Concrete Floor Systems 



15 



am 



Moment 




Shear 



p 



p 



p 





Loads at M/ddl e Points 



p p 



0.222 



P P 



OJII d33 




0144 




P P 



P P 



o.m 

fM flgfoirl AO &27W.27 




y. 



0.240 



Loads at Third Points 



p p p , 97 p p p 

I — £J 




p p p 




0.312 
P P P 




(A a -0.13 

im m 

p p p p p p p 

DM MWM I fimWi 



P P 




Loads at Middle and Quarter Points 

. i 



16 



Simplified Design of 



FORMULAS FOR REINFORCED CONCRETE DESIGN 

Standard Notation 



fa = 

fc = 
h - 

1/ = 

A. = 
A = 












= 



tensile unit stress in longitudinal reinforcement. 

compressive unit stress in extreme fiber of concrete in flexure. 

tensile unit stress in web reinforcement. 

modulus of elasticity of steel in tension or compression 

(= 30,000,000 p.s.i.). 
modulus of elasticity of concrete in compression. 

E c 

bending n inent or moment of resistance in general. 

effects area of reinforcement in tension in beams and slabs. 

f ff< ctive area of reinforcement in compression in beams and slabs. 

width of rectangular beam or width of flange of T-beam. 

depth from compression surface of beam or slab to center of 
longitudinal tensile reinforcement. 

depth from compressive surf ace of beam to center of longitudim 
compn ive reinfor ement 

w altli of web in T-beams 



/ = thickness of flang< in T-beams. 



3 



p = 
V = 

»v - 



ratio of depth of neutral axis to di i -i.l) d. 
!tjo lever arm of resisting couple to depth d, 

ratio of em ctive area of tensile rcinfora d nJ to effective uwh of 
ooncreb in beams or slabs = A x + bd 

of effective ana of compn iv< reinforcement to effective 

a lea of I ,ih. 

total shear. 

shearing unii 

shearing unit stress permitted « tie 

lue depi nding it« snchoragi 

foRM III 



>nrn itc of i h< web; D 

of the loi i udinftl nil 









h = 



/ 



(i 



u 
P 



- » 

!"'•"■ i group of i 

bond ui 

- ai ••|t:m l» jjf uf I.. r slab 

ctgofvert (irrupt, measured perpend ilar t h< 
tion of the stin 

istar <« u\onu )< of i,« : ( wl ch IH b r roifnui i- 

requir neaanred omth< •; oe .,f ^ipp' 
uniformly d U d J"»d j -tit of length of hear r slab. 



Concrete Floor Systems 



17 



1, Rectangular Beams and Slabs 



k 



.j, 



pn + (pn) 



2 _ 



pn = 



1 + 



/ 



s 



i = i 



k 

3 



nfc 



V 



A 



V* 



bd fsffe 



fc W, 



+ 1 



Irk 
2Js 



c 



fc 



ifcbkd 



b 




& ±n 



Fig. 2 



K = l Afckj or pfsj 



M = MckM* = Kbd* or bd* = 



2M _ M 



M = pfsM 2 or bd 2 = 



M 



Ao = 




fsjd 



OTf S = 



M 



A s jd 



U .X£« J*„*** 



k 



n(l - fc) kjbd* 



18 



Simplified Design of 



2. Rectangular Beams Reinforced for Compression 



k = 




Ml v + P'j ) + n\ V + p') 2 



H&d + 2p'nd'( k - % 



n(p + p') 



2 = 



fc2 + 2p'n[ fe - -j 



jd = d — z 



It d ' 



ifcbkd 



(n-!)f c p'bd [ 

Resultant 





Fig. 3 



/c = 



6J/ 



/wJ~3Jfc - 



W + 6 -^l fc 



A 




d 




h = 



.u 



j)jbd 2 



= nf t 



1 - k 
k 



k 



ft = nfe 



df_ 
d 



/.- 



.1/ 



^5 = ~7-, or p&d 



M = Kbd* 



.1/ = p'bd 



d 




M_ 
Kb 



V = 



7 



c 

■si 



Ui^Jli 



A' = 



/" A* *™ 

/c( 2 _ 6 + T 



-Vf « - 



*J * - 



^ 
d 



4-ffe- 



ri'V 



( onckete Floor Systems 



19 



J. T- Beams 

Case I. When the neutral axis lies in the flange, use the formulas for 

rectangular beams. 

Case II. When the neutral axis lies in the web. The amount of com- 
pression in the web is commonly small compared with that in 
the flange and is neglected in the following formulas: 



Jfc = 



1 + 



/ 



k = 



8 



nfi 



, f 



tV 



kd = 



2ndA s 4- bl 2 
2nA s + 2bt 



d — jd = 



3kd - 21 t 
2kd - t 3 



6 



<>;y • &+ 



/W 1 



■ 

3 = 



<7/ \2pn 



G 



«i 



orj = 



_ / / .'jfc - 2 </</ 



ifcbdxUl 




neutral 



axis 



Fig. 4 



/s = 



,1/ 



3/ 



^ *# t>jm 2 



/c = 



fsk 



n(l - *) 



M = IsA $ jd 



M - / 



"-B 



6/ X id 



4. Shear and Web Reinforcement 

V 

v = — - (for rectangular beams) 
5;d 



20 



Simplified Design of 



v = 



V 



b'jd 



(for T-beams) 



5. Bond 



V 



u — 



vb 



Zojd Zo 



or Xo — 



V 



ujd 



Illustrative Examples 

Computations for typical problems are given, using the tables in Section 
III and design data in this section as a basis, and following through to an 
estimate of the material quantities for each system. The computations are 
made by slide-rule. It is urged that the designer familiarize himself with 
each step, in order that he may learn the correct use of the tables. 

Problem 1 

Given an interior panel 18 ft. by 24 ft. with the following superimposed 
load: 

Partitions = 20 p.s.f. 
Floor Finish = 12 p.s.f. 
Live Load = 50 p.s.f. 

Total 



82 p.s.f. 




Column 









I 



I $ffm_ J 





Id'-O'c.c.Cols. 



Fig. 5 



If ceilings are plastered 
the weight of lath and 
plaster should be included 
in the total floor load. 

Design a solid one-way 
slab, using intermediate 
beams framing into gird- 
ers on column centers, as 
shown in Figure 5. 

Clear slab span equals 
8 ft. less width of beam, 
say 7 ft. Refer to Table 
No. 7. Down the column 
for spans of 7 ft. and op- 
posite the line denoting 
an interior span, it will 
be found that the nearest 
(heavier) load is 98 p.s.f. 
This requires a slab 3 in. 
thick. However, in order 
to provide adequate 



Concrete Floor Systems 



21 



space for conduits and reinforcement a 33^ in. thick slab will be used for 
which the allowable safe load is 167 p.s.f. 

Use Sy^-in. slab. This slab weighs 44 p.s.f. making a total dead and live 
load of 82 + 44 = 126 p.s.f. 

Then the area of steel required in a strip of slab one foot wide = 



126 



167 + 44 



X 0.23 sq. in. = 0.14 sq. in 



According to Table No. 6, '^s-in. rd. bars spaced S 1 /^ inches on centers 
satisfies this requirement, but for bond 2o must equal 



126 X H (8.00 - 0.84) 



- = 2.02 in. 



100 X Yh (3.50 - 0.75 - Y% X 0.375; 

Use %-in. rd. bars at 7 -in. ox* Bend alternate bars over the beams and 
continue them in the top of the slab for a distance of one-fourth of the 
next span length (or in this case 2 ft. beyond the center line of the beam) 
on each side, as shown in Figure 6. 

Reinforcement for temperature and shrinkage at right angles to main 
steel is also required. The spacing shall not exceed five times the depth 
of the slab or 18-in. The area of steel shall not be less than 0.002 bd with 
deformed bars or 0.0025 bd with plain bars. Use deformed bars. A s = 
0.002 X 3H X 12 = 0.084 sq. in. Use %-m. r d. bars spaced 15 in. o.c. 
The maximum allowable spacing is not exceeded. 



2'-D" 



8'-C"c.c. 



t ^4 dear rberrt bars 



• . f ■ c 



. c 



.- £'• 




Straight bars § clear [?> Jjf 



10" l-Q" 



4-2" 



2-0" 



'• i "JT.'. *-T-* 



o 



■ - & L- - - ■ • • - V ■ : \ 

^_ — * * * P - .- , j- - • . - * - . -.. , A 



1-6" 



,v - 



i ■ 

■ • *: 



- 



t;*; 




beam 



Fij:. 6 



Design of Beams 

Each beam carries a load of 8(82 + 44) = 1008 lb. per lin. ft. Clear 
span = say 17 ft. Referring to Table No. 16 (Safe load table of T-beams, 
interior spans), it will be found that a beam 10 in. by 18 in. is satisfactory. 
This beam can carry a load of 1200 lb. per lin. ft. 

Use 10-in. by 18-in. beam. The stem of the beam weighs lol lb. per lin 
ft., making a total load on the beam of (1008 + 151) = 1159 lb. per ft. 



*o.c. — spacing on centers of bars. 



22 



Simplified Design of 



As = 



1159 



X 1.46 sq. in. = 1.25 sq. in. 



(1200 + 151) 
Use one %-in. rd. bar, straight; two %~in. rd, bars, bent. 

The total end shear equals half the entire load carried by the beam on 
its clear span. Assuming the width of the girders to be 16 inches, the clear 
span of the beam is 16 ft. 8 in. 

Then V = l -^- (1159) = 9650 lb. and the unit end shear 



v = 



V 



9650 



= 69 p.s.i. 



bjd 10 X 0.875* X 16 

As this value exceeds the maximum allowable shearing unit stress of 
40 p.s.i., take care of the excess by means of web reinforcement, such as 
stirrups. For their design, refer to the "Diagram for Stirrup Spacing, 
page 69. 

en v 



M 




' - 69 



16.67 _ 29 _ n . * 
X — = 3.50 ft. By 



40 = 29 p.s.i. and v'b = 29 X 10 = 290. The distance a 

over which stirrups are required = — ( - , „ . . 

F 2 \v / 2 69 

code requirements, the maximum permissible spacing of stirrups = %d** 

= 0.75 X 16 = 12 in. 

Use %-in. rd, U-stirrups. Holding a straight edge on the chart for %- 
in. rd. U-stirrups, so that it intersects the vertical line at 290 and the 
horizontal line at 3.50 ft., the spacing is read off at the intersections of the 
straight edge and the heavy vertical lines. In this case there are no inter- 
sections with lines indicating a spacing of 12 or less, so that stirrups are 
used at the maximum spacing of 12 inches up to a point equal to a. In 
other words, the spacing is S — 12, 12, 12 in. Thus three stirrups are 
required at each end of the beam, or a total of six for each beam. 

The maximum bond stress should not exceed 100 p.s.i. 

vb 69 X 10 _ 
u = — • = — = 73 p.s.i. 



So 



2.36 X 4 



Design of Girder 

Clear span equals 24 ft. less width of column, assumed 16 in , or 22.67 ft 
The girder, in addition to its own weight which acts as a uniform 
load, must support two concentrated loads due to the read ions 



+ 



16.67 
2 



] 



X 151 = 9070 -I- 1260 = 10330 lb. or a total concent rated load 



at the third-points of the girder <=2X 10330 = 20660 lb 

• Sf r Table No 3 for CX&C4 value* of j. For practical purpose*; is aa- <sd a* % or 0.375 

••AU examples of *tirruj gn have been baaed upon a maximu iri£ «> I ■ <il 

lOUOUp.p.i. The ]fi • • J Cod*, a* t«ed aftt «34* ixamph?* v.- « pernutta maximum sparing 

id and a working atrca i I / f = 20000 p.a.j To < tf) p - "• rnply mk H 
■ •'• natt* m thf diagram* on pajr* ' - 



Concrete Floor Systems 



23 




Assume the weight of the girder = 375 lb. per lin. ft., uniformly dis- 
tributed. 

Referring to Table 1 and Fig. 1 of Moment Coefficients on Pages 14 and 
15 to be used in this case, the equations for maximum moments (both the 
negative moment at the column and the positive moment at the center of 
the span) are as follows: 

(1) -M = 0.211 X 20660 X 22.67 + — X 375 X 22.672 = 98,500 + 

JL £ 

16,000 = 114,500 ft. lb. = 1,375,000 in. lb. 

Assuming a width of girder stem of 16 in., the required effective depth = 

M_ = J l, 375,000 = J 54g = 23 5 / 24 in Qr j d h f 26 . ^ 

K*b T157 X 16 If 

Use 16-in. by 26-in. girder. 
A _ M 1,375,000 _ OQ 

S ~fsJd~ 20,000 X 0.875 X 24 " 3 ' 28 Sq - m * 

(2) +M = 0.122 X 20660 X 22.67 + ~ X 375 X 22.672 = 57^00 + 

12 

16,000 = 73,100 ft. lb. = 877,000 in. lb. 

Depth of compression flange, t = 3.5 in., divided by the effective depth, 

d = 24 in., equals - = 0.15, in which case the value of j is greater than 7 * 

d 

recommended for rectangular sections, but using j = ~ H gives conservative 

results for A s . 

, 877,000 onQ 

A s = - = 2.09 sq. m. 

5 20,000 X 0.875 X 24 4 

Use five %-in. rd. bars. Bend three bars over the supports and use one 

J^-in. rd. extra bar, 12 ft. long in the top of the girder at the columns. 

This gives an area of 3.24 sq. in. at this point, where 3.28 sq. in. is required, 

which is sufficiently close. 

22 67 
The maximum end shear, V = 20660 H — X 375 = 24,910 lb. 

04 Gin 

v = 77 ^77^r ^7 = 74 p.s.i. v' = 74 - 40 = 34 p.s.i. 

16 X 0.875 X 24 y P 

v'b = 34 X 16 = 544 

The maximum shear at the first intermediate beam = 20660 -f- 4.42 X 
375 = 22,320 lb. 

22,320 ft7 . 

v = —^ tt^z = 6/ p.s.i. 

16 X 0.875 X 24 F 

v' = 67 - 40 = 27 p.s.i. 
v'b = 27 X 16 = 432 



*See Table 3 for values of K, 



24 



Simplified Design of 



Use %-in. rd. U-stirrups. The maximum permissible spacing = Q.75d 
= 18 in. 

On the diagram of web reinforcement, page 69, place a mark at a height 
equal to 432 above a point 6.92* on the a axis. Place a straight edge to 
connect this point and a point 544 on the v'b axis, and read the stirrup 
spacing : *S = 9 at 6, 3 at 9 in. or, 24 stirrups per girder. 

The bond stress at first intermediate beam, u = — 77-^ — = 91 p > i. 



11 



* u j. 74 X 16 „„ 

At the support, u — — — = A) p.s.i. 

M 16.91 



The depth of the girder designed above may be reduced by providing 
compressive reinforcement in the bottom of the girder at the supports 
For this purpose advantage may be taken of the straighl bars used 

tensile reinforcement at the center of 1 he -pan by extending the bare beyond 
the face of the support far enough to develop the Stfl in bond 

A lime that two : ' ,-m. rd. bars are brought from i In- other side of the 

support to provide, with the straight bars in 1 he girder being designed, 

four 1 -in. rd bars, A' a — 1.76 sq. in. 

Determine the dept h of the girder and the tensile reinforcemen 1 required 
by using 1 coefficients given in i ible i, and the formulas: d = 



1 fsl 



A$ = pbd. The values nce< ? . to enter this table are: d La assumed to !<• 

1.70 



21 in., fit' - 2.0 in - = 10; 7/ - — = 

d ' b<1 16 X 21 



= 0()(i") J 



From the table interpolating: K ■= 201. p ■ 0.0115 



, / 1,375,000 of , - 01 . 

d = */_: ! — = 20 ,. sa 21 in. 

I -"I x 16 



1 - 0.0115 X 16 x 21 :; so in 



With this arrange 1 1 iett{ the depth may be reduced three inches hut the 
i« reinforcement must l>< increased about twenty per c< nt 



Material Quantities 



ConCTtU: Slfib — 

Beam! = 

Girder = 



1^ X 24 X 202 
X 101 X 10 07 
2.50 X 22 07 

Total 



120 I cu. ft. 
I 6 cu ft.. 
56.7 cu. ft. 

23:14 ML ft. 

8.64 cu yd 






».«•! \vm firnt ittUra 



I.. 1111 an J IN Hi. k|"i . ,ii 



: 



Concrete Floor Systems 



25 



Steel : Slabs = 3 X 



12 



3X 



2X7 

12 
2X7 



— X 16.67 X 8.0 X .376 = 



X 16.67 X 12.0 X .376 = 



12 



129 lb. (straight 
i bars) 

194 lb. (bent 

bars) 



22.67 X ^ X 19.5 X 3.75 = 133 lb. (temp. 

bars) 



15 



Beams = 3 X 1 X 18 X 1.502 

3 X 2 X 28 X 1.502 
3 X 6 X 3.58 X 0.376 



Girder = 2 X 24 X 1.502 

3 X 37.5 X 1.502 
1 X 12 X 2.044 
24 X 5.42 X 0.376 



81 lb. 
252 lb. 

24 1b. 
357 lb. 

Mr 



4561b. 

(straight bars) 
(bent bars) 
(stirrups) 



Total Steel 



72 lb. (straight bars) 
= 169 lb. (bent bars) 
— 25 lb. (extra top bars) 
= 49 lb. (stirrups) 
315 1b. 

= 1128 1b. 




id'-lrtcXois. 



Fig- 7 



26 



Simplified Design of 



Problem 2 

Given the same panel as in the preceding example, except that framing 
is as shown in Figure 7. 
Clear span of slab is approximately 8 ft. 
Total superimposed load = 82 p.s.f. 
From Table No. 7: Use syfain. slab. Weight = 44 p.s.f. 



A s = 



82 + 44 



X 0.23 = 0.18 sq. in. 2o = 



504 



100 X J4 X 256 



= 2.24 



117 + 44 
Use %-in. rd. bars at 6-in. o.c. 

Bend alternate bars over supporting beams. 

Temperature bars same as Problem 1; %-in. rd. bars at 15-in. o.c. 

Design of Beams 

Load = 9(82 + 44) = 1134 lb. per lin. ft. 
Clear span = 22.67 ft., say 23 ft. 

From Table No. 16: Use 12-in. by 22-in. beam. Web wt. = 232 lb. per 
lin. ft. 

1134 + 232 w oin „„ 
A * = 1150 + 232 * 2.18 -2.15»q.,n. 

Use five '''4-in. rd. bars, bend 3 bars in alternate beams and two bars in 
intermediate beams, thus providing five bars over all supports. 



V = (9 X 126 + 232) 



22.67 



= 15,500 lb. 



v = 



15,500 



12 X 0.875 X 20 
v* = 74 - 40 = 34 p.s.i. 

226 I X S = 5.20 ft. 



- = 74 p.s.i. 



a = 



2 



74 



v'b = 34 X 12 = 408 

Use "Y%-in. rd. U -stirrups. 

Spacing = 1 at 9; 1 at 12; 3 at 

15 in. 10 required. 



Check for bond u = = 75 p.s.i. 

11.80 

Design of Girders 

Concentrated load = 2 ( ^ [9(82 + 44)1 + ??^ X 232 J = 32,460 lb. 

Assume uniform load (girder weight) = 270 lb. per lin. ft. 
Clear span = 16.67 ft. 
Assume b' = 14 in. 

(1) -M = 0.119 X 32,460 X 16.67 + -L x 270 X 16.672 = 70,600 



ft. lb. = 847,000 in. lb 



12 



V 



847,000 



157 X 14 
Use lJf-in. by 2t~in. girder. 



= */385 = 19.6 (say 20 in.) 



Concrete Floor Systems 



27 



a++ k * a 847,000 Q . Q 

At the supports, A s = -r-^r^r tti^z — = 2.43 sq. in. 

FF 20,000 X 0.875 X 20 H 

(2) +M = 0.130 X 32,460 X 16.67 +— X 270 X 16.67 2 = 70,300 + 

1 - 

6,200 = 76,500 ft. lb. = 918,000 in. lb. 

A 918,000 OAO 

A* = ■ = 2.62 sq. in. 

s 20,000 X 0.875 X 20 * 

tj /Three %-in. rd. bars, straight. 
Two %-in. rd. bars, bent. 

Maximum End Shear = -^^ + ~^- X 270 = 16,230 + 2250 = 
18,480 lb. 

v = = 75 p.s.i. v = So p.s.i. 

14 X 0.875 X 20 F 

v'b = 35 X 14 = 490 

Shear at center = 16,230 lb. 

v = - = 66 p.s.i. v = 2b p.s.i. 

14 X 0.875 X 20 F 

v'b = 26 X 14 = 364 

Use %-in. rd. U-stirrups. Maximum spacing = 0.75 X 20 = 15.0 in. 
Spacing = 3 at 6, 9 at 9-in. 2Jf stirrups required. 

75 X 14 

Bond at support, u — - — —^— = 96 p.s.i. 

4 X 2.75 



Material Quantities 

Concrete: Slab = 18 X 24 X 0.292 = 126.1 cu. ft. 

Beams = 2 X 1.54 X 22.67= 69.8 cu. ft. 
Girder = 1.80 X 16.67 = 30.0 cu. ft. 

225.9 cu. ft. = 8.35 cu. yd. 

Steel: Slabs = 2 X ( -^\ X 22.67 X 9.0 X 0.376 = 153 lb 



2X6 



(straight bars) 



2 X ( — ^- ) X 22.67 X 13.5 X 0.376 = 230 lb. 



2X6 



(bent bars) 



12 

16.67 X — X 25.5 X 0,376 = 128 lb 

15 

(temp, bars) 

"511 lb 



28 



Simplified Design of 



Beams =2X2 X 24 X 1.502 = 

2 X214 X 37.25 X 1.502 = 
2 X 10 X 4.58 X 0.376 = 



144 lb. (straight bars) 
280 lb. (bent bars) 
34 lb. (stirrups) 



Girder = 





458 lb. 


3 X 18 X 1.502 


= 81 lb. (straight bars) 


2 X 28.25 X 2.044 


= 116 lb. (bent bars) 


24 X 4.75 X 0.376 


= 44 lb. (stirrups) 




241 lb. 


Total steel 


= 1210 lb. 



I 



*o 



- i 



\r 



Column 



v 



CI 



T ile&btn cJoist 
Slab 



j 

1 1 

i 
1 




18-Oc.c.Cols. 




$ 






¥ 




Fig. W 



Problem 3 

Given the same panel as in Problem 1, but framed as shown in Fig. 8. 
Design the panel, using one-way tile and concrete joist slab. Assume that 
beams are 16 in. wide. The clear span will be 16.67 ft., say 17 ft. The 
superimposed load is 82 p.s.f. 

From Table No. 8, Use6-in. tile and 8-in, top-ping, 6-in. joists - 18-in. o.c. 
The weight of this slab is 77 p.s.f. 



Concrete Floor Systems 



29 




A$ = = r—z — z=r-. X Steel Coefficient for — 

Load Coefficient 12 

82 4- 77 
= — t^k — X 1.05 — 0.50 sq. in. per joist 
332 J 

Us JOne H-in. sq. bar, straight) joigt> 
(One yi-m. sq. bar, bent J 

Bond at support: u = - = 71 p.s.i. 

Transverse reinforcement: A s = 0.002 X 3 X 12 = 0.07 sq. in. 
Use %-in. rd. bars at 15-in. o.c. 13 required. 
Stagger tile joints in adjacent rows of tile as shown. 

Design of Beams 

Load: 18(82 + 77) = 2860 lb. per lin. ft. 
Clear span: 22.67 ft. assuming 16-in. column. 
From Table 16: Use 16-in. by 30-in. beam. 

Web wt. = 16 X 21 X 150 + 2(112 - 77) = 408 lb. per ft. 

144 

. 2860 + 408 nQ 

As = r— r— X 4.08 = 3.67 sq. in. 

6 3230 + 408 4 

to jThree %-in. rd. bars, straight)^ = g 60 sq in 
(Three yg-m. rd. bars, bent ) 

V (2860 + 408) X ^ = 36,900 = ^^ — - = 94 p.s.i 

y 2 16 X 0.875 X 28 

v' = 94 - 40 = 54 p.s.i. v'b = 54 X 16 = 865 

a = ~ X ^-^ = 6.50 Max. spacing = 0.75 X 28 = 21 in 

94 2 

Use %-in. rd. U-stirrups. Spacing: 2 at 3, 4 at 6, 2 at 9, 2 at 12 in. 

$0 required. 

94 X 16 
Bond at support: u = = 92 p.s.i. 

tF 6 X 2.75 

Material Quantities 

Tile 16 X 16 = 256 units of 6 -in. Tile. 

Concrete 

ol , 18 X 24 X 9 256 X 6 1ftC n ,. 
Slab = 196.0 cu. ft. 

12 12 

Beam A X 22.67 = 52.9 cu. ft. 

144 

248.9 cu. ft. = 9.23 cu. yd. 



30 




Simplified Design of 


Steel 






Slab 


16 X 18 X 0.850 


= 245 lb. (straight) 




16 X 27.5 X 0.850 


= 374 lb. (bent) 




13 X 25.5 X 0.376 


= 124 lb. (temperature) 
743 lb. 


Beam 


3 X 24 X 2.044 


= 148 lb. (straight) 




3 X 37.8 X 2.044 


- 232 lb. (bent) 




20 X 6.08 X 0.376 


= 45 lb. (stirrups) 



425 1b 

Total Steel 1168 lb 



Problem 4 



Given the same panel as in Problem 1, framed as shown in Fig 9. Design 
panel, using one-way tile and concrete joist slab. 

The clear span of the slab, assuming 12-in. width of beams, is 11.0 ft 
The superimposed load is 82 p. si. 

From Table No. 8: Use Jf.-in. tile ■+• 2-in. topping, 5-4n. joist.s, 1? in. o.< 
Weight of slab = 51 p.s.f. 



A* = 



133 

309 



X 0.62 = 0.27 sq. in. per joist 



r (One 3 f in. rd. bar, straight) joist 
(One J/£-m. rd. bar, bent J 



Bond at support : u = - 



1040 



= 76 p.s.i. 



3.14 X K X 5 

Transverse or temperature reinforcement: A$ = 0.0025 X 2.0 X 12 = 
0.06 sq. in. I se }4-t'<- r( l- bars at 10 in. ox. 26 required. 
Stagger tile joints as shown in Fig. 9. 

Design of Beams 

Load: 12 X (82 + 51) = 1600 lb. per lin. ft. Assuming a 16-in. width 

of girder, the clear BpaD of the beam is 16.67 it., sa\ 17 ft. From Tall 

No. 10: Cse l.i-m. by in-,,,, beam, 

Wei, w\ = * U x 150 + 2(75 - 51) - 223 lb. per lin. ft. 

1 44 
A 1600 + 223 „ % M 

As = ]_ X 1 «*7 = 1.73 sq. in. 




I r 8i four %-in rd. bars, bend two. A s = 1.76 sq. in 



V = (1600+25 x 



16.67 



= 15.2001b 



v = 



I5.2<MI 



,' = si - 40 = 41 | 

41 16.07 

« = — X — = 4 23 ft. 



12X0 S75X1S 
b = 41 X 12 = 492 



= 81 j> s i 



Max i ing = 0.7'. X IS « 13 5 e 



Concrete Floor Systems 



31 



Use %-in. rd. U-stirrups, Spacing = 9, 12, 12, 12, 8 required per beam. 

81 X 12 
Bond at support: u = j , , __ = 103 p.s.i. (Sufficiently close to 



4 X 2.36 



allowable maximum) 




Column 



<J 



CM 



bearrij 



~i 






C> 



V^ 



1 



n. 



-i 



n 



i! 




LJ 



1 



n r-n n 



I I li 



r. 



^ 

s 



LJ 



I 



ste 

LjMLJltj! 

1 I I I 



HnH 



!l 




' J _Li_ J J 



rH 



Hn 



i 



<5> 



H n n 

Til i 



i 



k> 
J3j 



LJ 



!u 




LJ"lJ 



I I 



I I I — I 



V 



L 



i *n 



ia-0 m cx.Cols 



Fist. 9 



=1 






£3 



r 



* i 




J. 



Design of Girder 



Concentrated load: P = 2 



18 
2 




12 (82 + 51) 




+ 1 M 7 x 223 

£4 



350 lb. per lin. ft. 



32,500 lb. 

Assumed girder weight 

Clear span: 22.67 ft. 

The coefficients of wl 2 and PI for moment are obtained from Table No. 1 
and Fig. 1. 

At support: -M = 0.119 X 32,500 X 22.67 + -^ X 350 X 22.67 2 

= 87,800 + 15,000 = 102,800 ft. lb. = 1,230,000 in. lb. 



32 



Simplified Design of 




me b' — 14 in. 



v 



I se lJfri% by 26-in. girder. 



As = 



1,230,000 



s 



20,000 X 0.875 X 24 



= 2.93 sq. in 



At the center: 



1 



+.V = 0.130 X 32,500 X 22.67 + -=- X 350 X 22.672 

12 

= 95,800 + 15,000 = 110,800 ft. lb. = 1,330,000 in. lb 
Using d = 24 in. the required flange width is: 



b = 



1,330,000 



131.3 X 24 X 24 



= 17.6 in. Actual width is 34 in. 



A* = 



1,330,000 



= 3.16 sq. in 



20,000 X 0.875 X 24 
Use four 1-in. rd. bars, bend two. 

Shear at support: 

,. _ ^500 + 22.67 X 350 , b 



V = 



2 2 

20,210 



14 X 0.875 X 24 
v'b = 416 
Shear at center: 



= 69 p.s.i. 



v' = 69 - 40 = 29 p.s.i. 



V = 



16,250 



14 X 0.875 X 24 



= 55 p.s.i 



v' = 15 p.s.i 



v'b = 210 



Maximum stirrup spacing: % X 24 = 18 in. 

I '■' * s-nt. rd. U-siirrups. Spacing 5 at 9, 4 at 12, 2 at 15-in. 22 required 

Bond at end: U = ■ = 77 p.s.i. 

4 X 3-14 ' 



Ma terial Quant i 1 ies 

Tih 2X11X10 = 8W mite of 4-in. Tile. 



Cona Slab: (18 X 24 



6 
12 



220 X 



_4_ 
12 



Beams: 2 X 1.17 X I6i 

Girder: 1 94 X 22.07 



= 142.7 cu. ft. 

= 39.5 ru.fl. 

= 44.0 cu. ft. 

226.2 CU. ft, - 8.40 



cu 



Concrete Floor Systems 



33 



Steel Slab: 2 X 10 X 12 X 0.376 = 90 lb. (straight) 

2 X 10 X 18.3 X 0.668 = 244 lb. (bent) 
26 X 19 X 0.168 = _83 lb. (temperature) 

417 lb. 

Beams: 2 X 2 X 18 X 1.502 = 108 lb. (straight) 

2 X 2 X 28.1 X 1.502 =t 169 lb. (bent) 
2 X 8 X 4.1 X 0.376 = _25 lb. (stirrups) 

302 lb. 

Girder: 2 X 24 X 2.670 = 128 lb. (straight) 

2 X 37.6 X 2.670 = 200 lb. (bent) 

22 X 5.42 X 0.376 = 45 lb. (stirrups) 



373 lb. 
Total Steel = 1092 lb. 



Problem 5 




Given the same panel as in Problem 1, framed as shown in Fig. 10. 
ign panel using 20-in. metal pans and concrete joists. Use tapered 
pans. The clear span is approximately 17 ft., the superimposed load is 
82 p.s.f. 

From Table No. 10 use6-in. pans + 2-in. topping, 6-in. joists — 26-in. o.c. 
Weight of slab is 48 p.s.f. 

A s = — — — — X 1.27 = 0.68 sq. in. per joist. 

244 

Usei° ne %- in - Id - bar, straight) joist 
(One %-in. rd. bar, bent ) 

2370 
Bond at support: u = — = 83 p.s.i. 

4.72 X V 8 X 6.88 F 

Transverse reinforcement: The A. C. I. Code requires an area of at 
least 0.049 sq. in. per ft. width for floors carrying a live load not in excess 
of 50 p.s.f. Use %-in. rd. bars at 12-in. o.c. 17 required. 

Design of Beams 

Load: 18(82 + 48) = 2,340 lb. per lin. ft. The clear span is 22.67 ft., 
say 23 ft. 

From Table No. 16, Use ll^-in. by 28-in. beam. 

Beam Weight: ^—^ X 150 + 2(100 - 48) = 396 lb. per ft. 

144 

. 2340 + 396 QQ 

As = 2400 + 396 X 3 ' 31 = 323 Sq - m - 

T7 se JTwo 1-in. rd. bars, straight. 
Three ^s-in. rd. bars, bent. 



34 



IM1MJFI* l> I >KSH.\ 



V = 



J.S40 +396) X 2 ^- = 31,000 )l» 



2 



= 



1.000 



14 X 0.875 X 26 



97 p.s.i. 



r' = 97 - 40 = 57 p.s.i. 



v'b = 57X11 = 798 



5 j 



a = 



22 67 
i X ==: = 6.70 ft. 
7 2 



f v-/r. rd ll-strrrup* Maximum sp. iig = 19.5 in 

Spacing = ■"> at 6, 3 at 9, 1 at 15. 00 refftlWME. 

07 X 14 

Bond at support: u — — — ^— = 82 p s i. 



' 






: 



Column 




A 



ZTZ3 




.1 
3 



1 I 



1 



1 VfOi^etaJPin&CqncJoiji, 

■ - — -|j 



J/**' 



I 



till ^^^ 

i 
i 

H 



1 

"II 

- 



] 



i 






■n- 



I6 : 0"cc Cff/s 



I IK 10 



Mul *-r tnl ( hid n I 1 1 i#vs 



nrrrh 



IS > J I 



150.0 < fi 



Concrete Floor Systems 



35 



Beam: 1.94 X 22.67 = 44.2 cu. ft. 



194.2 cu. ft. == 7.20 cu. yd 



Steel 



Slab: 



12 X 24 
26 

12 X 24 
26 



X 18 X 1.043 = 208 lb. (straight) 
X 27.5 X 1.502 = 458 1b. (bent) 



17 X 25.0 X 0.167 



Beam: 2 X 24 X 2.670 

3 X 37.7 X 2.044 
20 X 5.75 X 0.376 



= 71 lb. (transverse) 
737 1b. 

= 128 lb. (straight) 

= 231 lb. (bent) 

= 43 lb. (stirrups) 
402 1b. 



Total Steel = 11891b. 



I 



^ 



* i 



i 




wf ~ 



Short Seam 



Two-Way 



\ 



Slab 



I 



Column 



i 



* + u 



18-0 c.c.Cols. 



Fig. 11 




6 










36 



Simplified Design of 



Problem 6* 

Given an interior panel, 18 ft. by 24 ft. between centerlines of beams as 
in Fig. 11, and a total superimposed load of 82 p.s.f. Determine size of 
slab and beams, amount of reinforcement, and quantities of materials 
required. 



Design of Slab 

Assuming beam widths equal to 1 ft., the ratio of long to short clear spans 

94. 1 

is — - = 1.35. For design of slab, use Table No. 19 (Ratio 1.4 to 1.0). 

18-1 

In column headed 23.8 X 17 and for Interior-Interior condition of con- 
tinuity, a 5)4-in. slab supporting 128 p.s.f. is seen to be sufficient. The 
5}^-in. slab weighs 69 p.s.f. 

Reinforcement given in table is: 

Short way: }^-'m. rd. bars at 5j^-in. o.c. (area per linear foot = 0.44) 

Long way: l Ar'm. rd. bars at 9-in. o.c. (area per linear foot = 0.27) 

With this reinforcement, the slab can carry 

128 + 69 = 197 p.s.f.; but the actual load is 82 + 69 = 151 p.s.f. 



Reduce the reinforcement above by multiplying it bj 



151 
197 



= 0.77: 



Short way (0.77 X 0.44 = 0.34) : >£-in. rd. bars at 7-in. o.c. 
Long way (0.77 X 0.27 = 0.21): J^-in. rd. bars at 12-in. o.c. 

This is the reinforcement required for center bands. The width of outer 

17 

bands, adjacent to beams, is — = 4.25 ft., and the width of center bands 

4 
is 23.0 - 2 X 4.25 = 14.5 ft. and 17.0 - 2 X 4.25 = 8.5 ft. Reinforce m-nt 
in outer bands is reduced to: 

Short way (0.75 X 0.77 X 0.44 = 0.26): V%m. rd. bars at 9-in. o.c. 

Long way (0.75 X 0.77 X 0.27 = 0.16): H-in. rd. bars at 14-in.** o.c. 

Total number of slab bars in a panel is: 



Short way: 
12 X 14.5 12 X 8.5 

T" 



7 
Long way: 



9 



= 24.9 + 11.3 = 36.2, say 36 barsi 1 ? f tra, 8 hl 

' J * 18 bent. 



12 X 8.5 12 X 8.5 

— ~r * — 



12 



14 



« 8.5 4- 7.3 = 15.8, say 15 bars 




straight 

bent. 



•Explanation and discussion pertaining to two-way concrete hlabs are given on Pages 4 B t->40. 
**)4m i* Ttiuxiriium Bpacing allowed. See explanation of Tables No. 17-22, on Pages 45 to 49. 



Concrete Floor Systems 



37 



Design of Beams 

Use beam load coefficients, (1 — er) and {1 — r) in Table No. 19 for 
Interior-Interior condition of continuity in the bottom lines: for bending, 
0.835 (long beams), 0.440 (short beams); for shear, 0.733 (long beams), 
0.267 (short beams). Column widths are taken as 16 in. 

Short Beam 

Clear span: 18.00 - 1.33 = 16.67 ft. Use Table No. 16, I = 17 ft. 
Superimposed load for bending: 

2X^X 0.440 X 151 = 1590 lb. per lin. ft. 
2 

Use beam 12 X 20 in. Weight of web: 250 - 69 = 181, say 180 lb. 

From Table No. 16: ^ s = L °™ , Z X 1.97 = 1.71 sq. in. 

1860 + 180 ' 

tj (Two ^-in. rd. bars, straight. 
Two Ji-in. rd. bars, bent. 

Superimposed load for shear: 

24 
2X^X 0.267 X 151 = 970 lb. per lin. ft. 
2 

V = (970 + 180) X i^ = 9,600 lb. 

9,600 

V = -• = Q-1 D.S.L 

12 X 0.875 X 18 P 

v' = 51 - 40 = 11 p.s.i. 
v'b m 11 X 12 = 132 

a = — X — *■ — = 1.8 ft. Maximum spacing = 0.75 X 18 = 13.5 in. 
51 2 

Use %~in. rd. U -stirrups. Spacing = 2 at 12 in. 4 required. 

12 
Bond at support: u — 51 X — = 65 p.s.i. 

^ 4 X 2.36 

Long Beam 

Clear Span: 24.00 - 1.33 = 22.67 ft. Use Table No. 16, I = 23 ft. 
Superimposed load for bending: 

1 o 

2 X — X 0.835 X 151 = 2270 lb. per lin. ft. 
2 

Use beam U X 28 in. Weight of web: 408 - 69 = 339, say, 340 lb. 

From Table No. 16: A s = ^ + |^ X 3.31 = 3.15 sq. in. 

2400 + 340 

Usefi™ *~* n ' rc *' k ars? straight. 
Two 1-in. rd. bars, bent. 



38 



Simplified Design of 



Superimposed load for shear: 

2 x — X 0.733 X 151 = 1990 lb. per lm. fl 
2 

V = (1990 + 340) X 2 ^- = 26,400 lb. 



v = 



26,400 



14 X 0.875 X 26 



2 

= 83 p.s.i. 



v' = S3 - -10 = 43 p.s.i. 
v'b = 13 X 14 = 600 
43 22 67 

a = — X r- = 5,87. Maximum spacing = 0.75 X 26 = 19.5 in. 



8 



2 



%4n rd. U-etirrups. Spacing 2 at <>. 2al <>, 12, 15, is m / ; required. 



Bond al support : u ~ 83 X 

M a t eria I Qua n 1 1 1 ies 

Com U 

Slab: 



14 



1 X 3.14 



= 93 



St< < I 



<\ d> 



Brain 



is X 24 X 0458 


= 


108.0 cu. f(. 




1 2i x i0.sa 


BX 


20.4 0U. fl. 




% L9 X 24.00 


- 


52.(1 CU ft. 








271.0 cu. fl ItiJ/ScU 


yd 


18 X 18.0 X 0.668 




217 11). (straight) 




is x 27.3 X 0.668 


= 


: i lb (brut) 




8 x 24 n x 668 


"^ 


128 lb. fraiulil) 




7 X 36.3 X 0.66S 


= 


170 lb. (bent) 




2 X 18.0 X 1 502 


m 


51 lb <slr:ii K lit) 




2 X 28.1 X 1.502 


m 


si lb (bent) 




J X 4 OS X 0.376 


= 


6 lb. (stirrups) 




I x 21 o x 2 676 


■ 


128 lb (straight) 




-'X 7 X 2 870 


■ 


201 lb (bent) 




M X 5 58 X 376 


C=: 


1 lb (stirrup 





Tntnl « l$J,r, (i 



^u m m<tr\ 



\ bi nhn< <y\ <i« :-i»j/i proeedurei I n pn nted in Proi > J t 

6 foi nsofc crete floor hi m I Y\w problem* also jjiv< <juar 

tit i which an helpful in making i mp« tm* \>ru toadopi 

ii i d' lx*st iuited to a v dH . 

i rotii tin -uMiinarv ifl Tabln 2, it )> < 'hat l<-f I b< fanl\ |0D| 

u i. and i li^bt l< I, diiio tl ing « uml, the ribbed Hooi 

I pai PquircH thr a*1 *!ightl> I >r< 



Concrete Floor Systems 



39 



TABLE No. 2— SUMMARY OF QUANTITIES 



Problem 
No. 



1 
2 
3 

4 
5 
6 



Type of Slab 



One-way Solid Slab (1) ...... . 

One-way Solid Slab (2) 

One-way Tile Filler Ribbed Slab (1). . 
One-way Tile Filler Ribbed Slab (2). . 
One-way Metal Pan Filler Ribbed Slab 
Two-way Solid Slab . 



Material Quantities 



Concrete 
Cu.Yd. 



8.64 
8.35 
9.23 
8.40 
7.20 
10.03 



Reinforce 

ment 

Lb, 



Filler Blocks 

No. and 

Size 



1128 
1210 
1168 
1092 
1139 
1346 



256 of 6" 
220 of 4" 



ment than the type requiring the least steel and is the lightest construc- 
tion because of the smaller quantity of concrete. This type of floor will 
in general prove most economical for the case considered. It can also be 
seen that the two-way solid slab is not well suited for this case. Had the 
superimposed load been much heavier, this type would be relatively more 
economical. A complete cost estimate must, of course, include the cost 
of forms. 

SECTION III— STRUCTURAL DESIGN TABLES 



TABLE No. 3— VALUES OF DESIGN CONSTANTS FOR VARIOUS 
COMBINATIONS OF STEEL AND CONCRETE STRESSES 



fs 










16,000 












11 
fc 

p 

k 

j 

K 






15 






12 


10 




1 650 

0.0077 

0,3786 

0.8738 

107.5 


700 

0.0087 

. 3962 

8679 

120.4 


750 
0.0097 

0.4128 

. 8624 

133.5 


800 

0107 

0.4286 

0.8571 

146.9 


900 
0.0129 
4576 

0.8475 
174.5 


1000 
0134 
4286 
8571 

183.7 


1125 
0.0161 
0.4576 
8475 

218.1 


1200 
0161 
0,4286 
8571 

220 4 


1350 
0.0193 
4576 
0.8475 

261.8 




h 










18,000 












n 

fc 

p 

k 

j 

K 






15 






1 


2 


10 




650 

0.0063 

0.3514 

0.8829 

100.8 


700 

C . 0072 

0.3684 

8772 

113.1 


750 

. 0080 

0.3846 

0.8718 

125.7 


800 

0.0089 

0.4000 

0.8667 

138.7 


900 

0.0107 

0.4288 

0.8571 

165.4 


1000 
0.0111 
4000 
8667 

173.3 


1125 
0.0134 
4286 
0.8571 

206 6 


1200 
0133 
4000 
8667 

208.0 


1350 
0161 
4286 
8571 

247.9 




fs 










20,000 












n 

fc 

p 

k 

* 

J 

K 






15 






1 


2 


10 




650 

0.0053 

0.3277 

0.8908 

94.87 


700 

. 0060 

0.3443 

8852 

106.7 


750 

0.0068 

. 3600 

. 8800 

118.8 


800 

0.0075 

0.3750 

. 8750 

131.3 


900 

0.0091 
0.4030 

0,8657 
157 .0 


1000 
0094 
0.3750 
0.8750 

164.1 


1125 
0113 
0.4030 
8657 

196.2 


1200 
0112 
3750 
8750 

196 9 


1350 
0.0136 
0.4030 
8657 

235 5 





40 



Simplified Design of 



TABLE No. 4— RECTANGULAR BEAMS WITH COMPRESSIVE 
REINFORCEMENT DESIGN VALUES OF p AND K FOR 

fs = 20,000, f c = 900, n = 1 5 



p' 


d'/d = 


= 0.02 


d'/d - 


= 0.04 


d'/d = 


= 0.06 


d'/d = 


= 0.08 


d'/d = 


= 0.10 


V 


K 


P 


K 


P 


K 


* 


K 


P 


K 


002 . . . 


0.0103 


180 


0.0102 


179 


0.0102 


Ml 


. 01 01 


176 


0.0100 


174 


004 . , . 


0.0115 


204 


0.0114 


200 


0.0112 


197 


0.0111 


194 


0.0110 


191 


006 . . . 


0.0127 


227 


0.0125 


222 


0.0123 


218 


0.0121 


213 


0.0119 


208 


008 . . . 


0.0139 


251 


0.0136 


244 


0.0134 


238 


0.0131 ' 


231 


0.0129 


225 


0,010 , . . 


0.0151 


274 


0.0148 


266 


0.0145 


258 


. 01 41 


250 


0.0138 


242 


0.012 . . . 


0.0163 


298 


. 01 59 


288 


0,0155 


278 


0.0151 


268 


. 01 48 


259 ' 


0.014 . . . 


0.0175 


321 


0.0170 


310 


0.0166 


298 


0.0161 


287 


0.0157 


276 


, 016 . . . 


0.0187 


345 


0.0181 


331 


0.0177 


319 


0.0172 


306 


0.0167 


293 


0.018 . . . 


0.0199 


368 


0.0193 


353 


0.0188 


339 


0.0182 


324 


0.0176 


310 


0.020 . . . 


. 021 1 


392 


. 0204 


375 


0.0198 


359 


0.0192 


343 


0.0186 


328 


P 


d'/d = 


= 0.12 


d'jd = 


= 0,14 


d'/d = 


- 0.16 


d'/d = 


= 0.18 


d'/d = 


= 0.20 




















I 




V 


K 


P 


A' 


V 


K 


V 


K 


P 


K 


002 . . . 


0100 


173 


. 0099 


171 


. 0099 


170 


. 0098 


168 


. 0097 


167 


004 . . . 


0.0109 


188 


0.0107 


185 


0.0106 


182 


. 01 05 


180 


0.0104 


177 


006 . . . 


0117 


204 


0.0116 


199 


0.0114 


195 


0.0112 


191 


. 01 1 


187 


008 . . . 


0.0126 


219 


0.0124 


214 


0.0121 


208 


0.0119 


203 


0.0116 


198 


010 . . . 


0.0135 


235 


0.0132 


228 


0.0129 


221 


0.0126 


214 | 


. 01 23 


208 


012 . . . 


. 01 44 


250 


0.0140 


242 


0.0136 


234 


0.0133 


226 


0.0129 


218 


014 . . . 


0.0153 


266 


0.0148 


256 


0.0144 


246 


0.0140 


237 


0.0135 


228 


016 . . . 


01 61 


282 


01 57 


270 


. 01 52 


259 


0.0147 


249 


0.0142 


238 


, 018 . . . 


00170 


297 


0.0165 


284 


0.0159 


272 


0,0154 


260 


0.0148 


248 


0.020 . . . 


0.0179 


313 


. 01 73 


298 


0.0167 


285 


0.0161 


271 


0.0154 


259 



TABLE No. 5— SECTIONAL AREA IN SQ. IN. 
OF VARIOUS NUMBERS OF BARS 



Number 


Size . . 


W4 


H m + 


M'° 


W* | 


W4> 


W* 


i% 


■j*a 


1H' a 


VM ° 
























of 


Wl. per Ft. 


0.376 


0.668 


0.850 


1.043 


1.502 


2.044 


2.670 


3.400 


4.303 


5.313 


Bars 
























Perimeter , 


1.18 


1.57 


2.00 


1,96 


2,36 


2.75 


3.14 


4.00 


4.50 


5.00 


1 


0.11 


0.20 


0.25 


0.31 


0.44 


0.60 


0.79 


1.00 


1.27 


1.56 


2 


22 


. 40 


0.50 


0.62 


0.88 


1.20 


1,58 


2.00 


2.54 


3 12 


3 


0.33 


0.60 


0.75 


0.93 


1.32 


1.80 


2.37 


3.00 


3.81 


4.68 


4 


0.44 


0.80 


1.00 


1.24 


1.76 


2.40 


3.16 


4.00 


5.08 


6.24 


5 


0.55 


1.00 


1.25 


1.55 


2.20 


3.00 


3.95 


5.00 


6.35 


7.80 


6 


0.66 


1,20 


1.50 


1.86 


2.64 


3.60 


4.74 


6.00 


7.62 


9,36 


7 


0.77 


1.40 


1.75 


2.17 


3 08 


4.20 


5.53 


7.00 


8.89 


10.92 


8 


0.88 


1.60 


2.00 


2.48 


3.52 


4.80 


6 32 


8 00 


10,16 


12.48 


9 


0.99 


1.80 


2.25 


2.79 


3.96 


5.40 


, 7.11 


1 9 00 


11 .43 


14 04 


10 


1.10 


2.00 


2.50 


3.10 


4.40 


6.00 


7.90 


1000 


12.70 


15.60 


11 


1.21 


2.20 


2.75 


3 41 


4.84 


6.60 


8.69 


11.00 


13,97 


17,16 


12 


1 32 


2.40 


3.00 


3.72 


5.28 


7.20 


9.48 


12.00 


15.24 


18.72 


1 13 . . . 


1 43 


2.60 


3 25 


4.03 


5.72 


7.80 


10.27 


13.00 


16.51 


20 28 


14 


1.54 


2 80 


3,50 


4 34 


6.16 


8.40 


11.06 


14.00 


17.78 


21 84 


15 


1.65 


3.00 


3.75 


4 65 


6.60 


9.00 


11.85 


15.00 


19.05 


23.40 


, 














' 







Concrete Floor Systems 



41 



TABLE No. 6 



AREA OF STEEL IN SQ. IN. PER LINEAR FOOT 
OF SLAB FOR RODS SPACED AT 

VARIOUS INTERVALS 



Spacing c. to c. in Inches 


Wi> 


w+ 


y 2 >o 


«** 


?r0 


W + 


1*0 


T° 


m* a 


1J4' D 


3 

Wi 

4. 

4H 


0.44 
0.38 
33 
29 


0.80 

0.69 

0.60 
0.53 


1.00 
86 
0.75 
0.67 


1 24 
1 06 
93 
83 


1 76 
1.51 
1 32 
1.17 


2 40 
2.06 
1.80 
1.60 


3.16 
2 71 
2.37 

2 11 


4 00 

3.43 

I 3 00 

2 67 


5.08 
4 35 
3.81 
3.39 


5 35 
4 68 
4.16 


5 

5^ 

6 

6H 


26 
24 
22 
0.20 


0.48 
44 
40 
0.37 


0.60 
0.55 

50 
0.46 


74 
68 
62 
57 


1 06 
0.96 
68 

81 


1.44 
1 31 
1 20 
1.11 


1 90 
1 72 

1.58 
1.46 


2.40 
2 18 
2.00 
1 . 85 


3 05 
2.77 
2 54 
2.34 


3 74 
3.40 
3 12 
2 88 


7 

7K 

8 

BH 


0.19 
18 
0.16 
0.15 


. 34 
0.32 
0.30 
0.28 


43 
40 
37 
035 


53 
0.50 
0.46 
44 


75 

70 
66 
62 


1 03 
0.96 
90 
85 


1 35 
1 26 
1 18 
1 12 


1 71 

1 60 

1 50 

, 1.41 


2 18 
2 03 

1.90 
1 79 


2 67 
2 50 
2.34 

2 20 


9 

9H 

10 




27 
25 
0.24 


0.33 
32 
30 


41 
39 
37 


59 
56 
53 


80 
76 
0.72 


1 05 
1 00 
95 


1 33 

1 26 
1.20 


1 69 

1.60 
1,52 


2 08 

1 97 
1 87 



EXPLANATION OF TABLE No. 7 

To determine the depth of slab for a given superimposed loud, clear span 
and "condition of continuity/' read down the column for the given clear 
span and across from the given "condition of continuity" until a safe load 



At. 



+<\ 



Simple Span 




End Span 



ii 



it 



5 



I 



ll 



<L 



<L 



Interior Span 



ti 



4 



il 



s 



\- 



Fi B . 12 



I 



s 



il 



4 L 



42 



Simplified Design of 



is reached, which is equal to or greater than that required. The thickness of 
slab is shown in the first column of the table. 

If the safe load is equal to the given load, use the size and number of 
bars shown in the fifth and sixth columns. If the safe load is larger than 
the given load, the steel area shown in the fourth column may be reduced 
by the proportion: 

given superimposed load + weight of slab 

safe superimposed load + weight of slab 

The size and spacing of reinforcing steel may then be selected by use of 
Table 6, page 41. Problem 1, page 20, illustrates the use of Table 7. 

Typical bar bending diagrams for simple, end and interior solid slab 
spans are shown in Fig. 12. 



EXPLANATION OF TABLES No. 8 TO 12 

The lables of safe loads for ribbed slabs are based upon clear spans, 
assumed 1-ft. less than the center to center spans, for both moment and 
shear. In the tables for floors involving tile fillers, 1^-in. has been added 
to the joist widths when computing the shearing value of the joist. This 
is done in order to take into account the shells of the tile in contact with 
the joist. Such allowance can only be made when the joints in the tile are 
Staggered.* 

The safe superimposed loads in Table 8 have been computed, taking 
into consideration the following weights of filler units (each unit 12-in. by 
12-in. by the depth shown): 



Depth of 


Weight per 


Tile 


Unit 


In. 


Lbs. 


4 


16 


6 


22 


s 


30 


10 


36 


12 


40 



If the weight of filler units used varus from thai in the above tahl< 

the tabular values of safe supen m ]>( i Sid IsmtiBl be corrected accordinglj 

In order to design a ribbed Blab having the clear span and tike super- 
imposed load in pounds per sq. ft. enter the proper table (depending on 
type oi filler units) in the column for the gives span; follow down the 
column until a safe load is reached equal or greater than the given load; 
read tin -lab dimensions and joist width in columns one and three. Tin 
I of steel required per joist is determined hv the following formula: 



A* m 



Design Load + Wt ight of Slab 

Load Coefficient 



X Steel Coefficient 



* ' ' ' Khf * ! I !'■ ar< coiuridi l effei i u oomprattioo inee for oogitivi mend near vuppori 

surrj«-d in »j >i*t *l*i - that naprewve *1 I at the interior nupj i of two »p*nr it i qi 

J hi rmdjrpftra for positive n i at and &i interior »uj ' three or in i ».» 

P r < miti tforcemenl The Ic-nj* 4 ...hnh, rsigl 

fhhW U ucr< )<•*>> \huu 20 cJ beyond end of il in1 r • 



Concrete Floor Systems 



43 



The Load Coefficient is found in the table just above the safe load value for 
the particular span and slab dimensions. The Steel Coefficient is found 
opposite the safe load value in columns 6, 7 or 8, depending upon the 
condition of continuity. 



Illustrative Problem 

The following example illustrates the use of fables 8 to 12, inclusive. 

Design a tile and concrete joist slab, interior panel, having a clear span 
face to face of supporting beams of 16-ft. and carrying a superimposed 
load of 100 p.s.f. Enter Table 8, under span lb-ft., follow down the column 
to the first safe load value greater than 100 p.s.f.; namely, 107 p.s.f. In 
the first column it is found that an "8 -f- 2 Slab" is adequate using a 5-in. 
joist as indicated in Column S. The weight pi the slab, including the tile 
filler, is 75 p.s.f. 

The load coefficient is 416 and the steel coefficient for the condition of 

continuity, — is 0.95. The steel area required is then obtained from the 

12 ' 

formula for An given above. 



As = 



N 



(100 + 75) 
416 



X 0.95 = 0.40 aq. in 



Table 5 shows two j^-ln. rd. bars have an area of 0.40 aq. in. 
Typical bar bending diagrams for tile and metal pan filler and concrete 
joist slabs are shown in Fig. l.'i. 



1 



ii 



H 



- 



1 



il 



I 



t 



^ 



■A 



^ 



^ 



il 



T 



I 



End Spa n 



,\ 



ii 



1 



z 



AT. 



X_ 



7^ 



\ 



il 



Inter /or Span 



I 



f 



I 



t 



>1 



- 



L 



f 



1 



Via. 13 



44 



Siiuplified Design of 



EXPLANATION OF TABLES No. 13 TO 16 

Table No. 13 is a safe load table for beams of simple span designed 
for a moment of M = -\-}^wl 2 at the center of the span. Tables No. 14, 
15 and 16 are for beams designed for moments at the support of — }/%wl 2 , 
— 1 fXOwl 2 and — l/\2wl 2 , respectively, and differ from the preceding tables 
in that the unit stress in the concrete, /c, is taken as 900 p.s.i., instead of 
800 p.s.i., as provided in the A. C. I. Code for compression in concrete 
at beam supports. Coefficients of id 2 for various span arrangements are 
given in Table 1, page 14. Typical details for bending bars are given in 
Fig. 14. 




End Span 




Interior. Span 

Fig. 14 



Illustrative Problem 

D< sign a pimp]*- i in with a char span of 22 ft. The beam carri* a 

superimposed load, excluding the weight of the beam, of 12(H> lb. per Lin, 
ft. and tli< depth of the slab (thickm as of flange) is 4 in. 

Enter Tabl< No. 13 and folio* down the column for ' i;ui - 22 ft." 
until a safe superimposed load greater than 1200 pl.f. is found opposit< 

a flanp thickness of 4 in., namely 1230 pl.f. The depth of I will b< 

4 in. given in Column 1 and the width of nreb wrill be 12 in., shown in 

Column 3. The m ight of tJi< nreb ifl 125 p.l.f 

The an a oi 3teel required for a superimposed J<<;<d of 1230 lb. per hn. 
ft is 4.68 sq. in. as found in Comma 7. This may !><• reduced in proporl n 
to the actual 1 d load on tl • beam. 



Concrete Floor Systems 



45 



A AfLQs, 1200 + 125 A CO 

As = 4 - 68 X 1230 + 125 = 458 Sq - m ' 

According to Table No. 5, this area can be obtained with three lK-in. 
sq. bars. (A s = 4.68 sq. in.) 

If the span had been 20 ft. the dimensions of the beam would be gov- 
erned by shear and the area of steel would be determined by the design 
coefficients. The same dimensions would be required as in the previous 
case, the safe superimposed load being 1390 lb. per lin. ft. The steel coeffi- 
cient, cu is 0.216 and A s = c t l = 0.216 X 20 = 4.32 sq. in. This area can 
be obtained with three 1-in. rd, bars and two 1-in. sq. bars. A s = 4.37. 

The bond stress must be checked and shear reinforcement provided if 
necessary to complete the beam design. See Problems 2 to 6 for the method 
to be used. 



EXPLANATION OF TABLES No. 17 TO 22 

The use of spans equal to clear distance between beam webs is specified 
in some building codes; in other codes, longer slab spans may be specified. 
In either case, the designer enters Tables No. 17 to 22 with values of 
Spans in Feet computed according to the code that governs; the use of 
the table is the same in either case. 

The two-way concrete slabs in Tables No. 17 to 22 are designed accord- 
ing to the A. C. I. Code. The reinforcement given is that which is required 
in slab zones with maximum reinforcement, the balance of the reinforce- 
ment being determined as follows: 

(1) Positive reinforcement adjacent to a continuous edge only and for 
width not exceeding one-fourth of the shorter dimension of the panel may 
be reduced 25 per cent. 

(2) At a non-continuous edge, negative reinforcement per unit width, 
in amount at least as great as one-half that required for maximum positive 
moment for the center one-half of the panel, shall be provided across the 

a entire width of the exterior support. 

(3) The spacing of reinforcement shall be at most 23^ times the slab 
thickness and the ratio of reinforcement shall be at least 0.003. 

Reinforcement in the long direction of the slab has been assumed to be 
in the top layer; some codes call for this reinforcement to be in the bottom 
layer. Economically, there is little difference between the two schemes. 

Explanatory remarks, in addition to those in Problem 6, will be presented 
in the illustrative problem which follows. An end panel will be designed and 
the slab reinforcement detailed. The distribution of slab load to beams will 
also be discussed. 



4(i 



SiMi'Lii ii i> I >i >i».n <^ 



Illustrative Problem 

i u ili. [ in Fig 15 end span is in th< iort din ion and interim 

span to the long dii listam between ©ente rime ms being 

18 ft and 24 ft Let the Buperim posted load l» th< m - m 

Problem 6 S2 | \>>uim a t .m width of 12 m for interior abort b imi 

H in f.,j n ir long l»« am and 10 in. for BXU >r brain 1 ». i. rttiinc tl 

slah thiekn< number, si* and length of bat> required 
//, The ratio of lonj short clear span is 



24 <> (05 j (I "i . n 
|s<» - id ti H- 0.4) IT 



- 1 



r design of slab, use Table No la Ratio 1 1 (<• 1 o In tumn headed 

; s , / ;i n<j for Eud-lnU o condition of oontinuitj s6-in slabcnpabli- 
tip). i -i, imp( : I- ■ i "i MH | t to be pull of 

Tin ''in slab ti- 7"» | 

I;, ti i in (able it; 



Short m ' ti rd bars at 4-in. o.c (area | linear foot ■ O.ttO 
I. mi. : ' 2-in rd ban it G-h ■ > (a iPi lini u II 10 

I'lii b-in --lull with ili ■ nforcemenl i in rarrj Rafelj .i total I d of 
!i (i |, inn ib si luaJ ad ii onl ' l . | f, 

1 undei (he Codi pi lireinenU use a slab thinner 

ii- i ■ • r i . lit j * n i ■* | li Tin reinfoi Nov i, m:i\ I 

• ! in i In ion — 0(17 but it shall l.< riot ban 1 1 in 

n | 'lii. r minimum b 



I ! 






V II > \'J / Mi I 



III" j I 



! 



(I i |2 X 6 I ] r n 17 ^\ m 

if III I I. 19 u 

! a I 



i 



i 



, l - (I \K\ 1 

n . in - o . 



in i 



t ft 






• < 



Jut* *1 ih' 1 1 tii < r lmn<l 

\ »• i . I< I Jiilmji, 1 nf« 



w 1 









t I 



(I ( 



m » 



< 0.60 - ( 1 

n in - i i 



I I 



I ml 



< 



All 



;■ , . . 






- 



Concrete Floor Systems 



47 



« 



Detailing 

For the slab designed above, placing plan, bar schedule and bending 
details are shown in Fig. 15. According to (1), the width of all outer bands, 
b and e, is 34 X 17.0 = 4.25, where 17.0 is the shorter panel dimension. 

The bars in band a are J^-in. rd. at 6-in. o.c, the band width 

23.0 - 2 X 4.25 = 14.5 ft., and the number of bars * = 29, of 

6 

which 15 are straight and 14 are bent. For two bands 6, the number of bars 

12 X 2 X 4 25 
is : — = 12, of which 6 are straight and 6 are bent. 

o 

For the outer band, e, in the long direction, the bars are %-\n. rd. at 

12 X 4 25 
12-in. o.c, and the number of bars is — — — = 4, of which 2 arc straight 

1 W 

and 2 are bent. The remainder of the short panel width, 3 4 X 17.0, has 
^2-in. rd. bars spaced 9-in. o.c, and the number of bars in this band, rf, is 

'— — = 17, of which 9 are straight and 8 are bent. 



h 






m 



I 



I 



=t#= 



6-0" 



b 



24- 



c (top bars) 



u 



jfe— !__. -: 

— ( ^ 



b 



rE 



P=t 



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Mark rn3 



X 



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6-0" 



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Band 


No. 


Type 


Size 


Length 


Mark 


Spacing 


9 


IS 
14 


sir. 

bt 


i* 


16-0" 
23- &■ 


ml 


Q> "o. c. 


b 


3 
3 


str. 

bt. 


i* 


18-0" 
25-6' 


ml 


Q- 'o.c 


c 


18 


bt. 


i*\ 


5-3" 


m2 


IS'o. c. 


d 


9 

6 


str. 

bt 


i* 


24-0" 
36-3' 


m 3 


3' o.c. 


e 


2 
2 


5tr. 
bt. 


i* 


24-0" 
36-3' 


m 3 


12" ox 



Note; Bars in a and b in bottom layer 
Mark m2 



t A ., 



5-0 



—£dge of slab 



Mark ml 




Fig. 15— PART FRAMING PLAN AND BAR SCHEDULE 



Top bars in band c shall, according to code requirement, have an area 
per linear foot equal to one-half of that of bars in band a; that is, the 
reinforcement in c shall be J^-in. rd. bars at 12-in. o.c, or equal. The bent 
bars in 6, J^-in. rd. at 16-in. are deficient in area, and all bent bars are bent 
up too close to the exterior beam to be fully effective as top bars. The 
deficiency will be made up by using additional top bars ihrin. rd. at 15-in. 

1 2 X 23 
o.c, the total number in c being = 18 bars, which are hooked. 

15 



JS 



[MPLIKIKD 1 SIGN OF 



Distribution of Load from 
Ttco-U"ay Slabs to Beams 

The load transmitted from a two-way solid slab to the supporting bean 
according to tin \. C I. Code, may be determined as follow ! el 

H = Slab reaction on beam in lb. per Lin. ft. 
U> = Total slab load in p.s.f. 
L = Longer span in ft. 
/ = Shorter span in ft. 



Tin i m on the beams is then : 



wl 



On longer beam:/?£ ■» (I — er) X — 



tid 



1/ - X- 



« )n shorter I tm : A'/ =('/-■ X 



u>L 



an. I (/ - r) X 



ujJL 
2 



The (piant n ies {1 — er) and (I - / are used fco denote i oefficicnl which 
function of L I and tht condition of continuity rfcw meaning an< 

ppln-al inn of lb--, quantities will l» I xplained ami illi ted m roiun-e- 

tion w ii b I iu' 16, which refers 1 i two-way si iswithraf otLilnl 1:1.0 

■ ' L6 show- mm' groups Willi from one Vo hum- lI ii each group 
inrkid from 1 to (9 Nott thai group 8-3 applies tO Boc>r> having 

thn "f spans in either direction. The numb rs in parent hem < i>" 

the type.* of "< ondition of Continuity descril I m the four! 

ii* I fifth I tun ns of the two-w lab tabl< - Refer, for < sample, to I able 

V 19, in which th< condition of continuity □ the first line for parh **l:«b 
K Short Way Simple 11 and "Loup V\a> - i pi I on ng 

ii- fart< applyti bonding in brains, the fi t lit- in tin Im a! 

- ride of Tabli N 19 jnves — for tin* condition of ^nuplt -Simpl< 
tin* two \aln r (/ — ei (J i ffor bug beam and M€ for -hori 
beam) The a ut; | continuii Sim pi Simple, iimI the numben 

in 1 jn iwn if) hlali I adjacent to the U anr to \\ I 

ppl) Kin ti! rh of the nine lino of two mimU r> in tin- fx'\<5i a 

>p ptioi f t itinuity in th< fourth and hlih 

luu n in r) h< u ntal lii * n in 1 ig. M II 

ilar v Ik >k< lied ration oth« f than I 1 i n an« 

• vahe / — - and 1 — r ale d I III i chart 



illustrs of I m l» - again \ \\i I .'> and assum 

the ulah pI a i» part of i fi ftnng | h as I ig 16, 1 

I 15 *lal< ma d ' hi I ijc M> rh« stare- betw-c ; 

n slii •■ an. '< • ]Kft the sh way at 24 f 



Concrete Floor Systems 



49 




Sim. ' Simple spa n End* End span Int. a Inferior spsn 




1-2 





! 



0.555 
EnJ[(5) 



? 






* 



0.63 
Til 



§ 

V 



End 



(5) 



a 

0.835 



^ 
^ 
> 
^ 



063$ 
End\r$)_ 



^ 



^ 
S 



0.335 



7UJT 



^ 
> 
> 
^ 



End 



(5) 



0.335 



i 



^ 
^ 

> 

^ 



0335 



End 



% 



C£L 






Mil 



S 



0.335 
I 



A? 



I 



•5 



m 



335 



0.775 



3 






£221 



0.775 






Inf. 



*S 



f&) 



^ 

fc 
** 



01 IS 



om 



i 



SJ3L 



0.335 



1 



End 



% 



cs) 






63S 



2-2 



Z^A 




AS; 


is 


^ 

^ 
> 
^ 


s 

^ 
^ 


AS 

End 






3 ^ 

06 


35 


0.6 


35 


oSl 

£f fntf 








d.6 

inch 

05 


44 

\(B) 

£0 


<5£ 


* 3 

£<?<! 


?0 


OS 35 

y EnAfsl 


5 

>* 


> 


9A 

End 


|4 




0.5: 


?5 


3 


1 

35 



^ 

>* 

M- 
^ 



0.335 



End 



I 



(5) 




0775 
Int\fc) 



0.435 



IS. 



0330 
Eng\(6) 



ft 



■n 



0.330 






End 



X 



(5) 






1 0. 335 



I 



>> 



^ <s 



775 



^ 
> 

> 

^ 



IntSf9) 



^ 

>* 



> =* 



335 
End\{5) 



i 






Mil 



0.330 



0.835 






0,775 



I 
^ 



Int. 



^^ 



% 



(6) 



0-775 






End 






w 






0.880 



im 






End 






(5) 






Mil 



3-2 



3-3 



Fig. 16— FACTORS FOR DISTRIBUTION OF SLAB LOAD TO BEAMS 
FOR RATIO OF LONG TO SHORT SPAN = 1.4:1.0 FOR CALCU- 
LATION OF BEAM MOMENTS. SEE TABLE 19 FOR SIMILAR 
FACTORS FOR DETERMINATION OF BEAM SHEARS. 

the long way. Determine equivalent loads to be used in calculating moments 
in beams along the four sides of slab (6), the total load being 

82 + 69 = 151 p.s.f. on slab (9) 
82 + 75 = 157 p.si. on slab (6) 
82 + 81 = 163 p.s.f. on slab (5) 

The reaction on the beams in lb. per lin. ft. is as follows, exclusive of 
the weight of beam webs and using the formula given above with the 
value of (1 — er) taken from slabs (5), (6) and (9) in Fig. 16 or from lines 
5, 6 and 9 in the box in Table No. 19 : 

18 
Spandrel Beam: 0775 X 157 X — 



= 1100 



-I Q 1 C 

Long Int. Beam: 0.775 X 157 X ^ + 0.835 X 151 X — = 2230 

24 24 

Short Int. Beams: 0.540 X 157 X -£ + 0.440 X 163 X — = 1880 

These loads plus the weight of beam webs are to be used for calculating 
beam moments. For beam shears, use the same procedure but substitute 
values of (1 — r) for (1 — er). 



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shown in the table. 
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the heavy line are 
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ta ui oiugi iiiuoi 

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TO USE THIS CHART 

V 
Find fhe unif shedr t y *jjp 

Then /'= v- v c dnd dis is nee 3, over 

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{({) Compute v'b PI see a 

straight-edge on the diagram 

connecting the calculated 

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where their respective heavy 

lines cross the sfraight-edge 

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spacing permit fed , until 

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V '4 0,5 00*, the beam having 

3 uniformly distributed load. 

Required the number a nd 
spacing of i "t IT S tirrups in 
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r- 110-40 = 10 ps.i 
v'b- 14x10 -360 

Max allowable spacing = 
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s=&';G;3e3;i2;4t2i5" 

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81 



82 



Simplified Design of 



SECTION IV— SPECIAL DETAILS OF 

CONSTRUCTION 

Openings in Slabs 

Openings of various sizes and shapes will be encountered in any floor 
slab, due to stairs, elevator shafts, pipe shafts, and ventilating ducts. The 
method of providing for these depends largely on the ingenuity of the 
designer. Effort should always be made to preserve the continuity of the 
floor as far as possible and to eliminate framing beams or girders if it can 
be done without sacrificing stiffness and safety as a whole. 

Small openings usually require merely a rearrangement of the reinforcing 
steel and the introduction of a few additional bars. A right and a wrong 
method are illustrated in Fig. 18. 



Wall ^ 




W3I! 



2 



Beam 




WRONG METHOD 



Beam 



CORRECT METHOD 



Fig. 18 



Larger openings may sometimes be framed in one-way ribbed slabs by 
introducing header joists at right angles to the main joists and framing 
into wider or double joists along the sides of the opening. 

Very large openings may be more economically framed by marginal 
beams, which in turn frame into the main supporting beams or girders. 



Stairs and Stair Platforms 

Reinforced concrete stairs usually consist of an inclined slab with the 
steps formed on its upper surface. These may be quite simply designed b 
reference to the following table. 



Concrete Floor Systems 


83 


TABLE No. 23— TABLE OF CONCRETE STAIR SLABS 




f s = 20,000 p.s.i. 1" Clear Protective 




f c = 800 p.s.i. Covering 




n = 15 Plastered Soffit 




L.L = 100 p.s.f. 




Horizontal Span of Stairs 
in Feet 


Total Thickness of Slab 
in Inches 


Reinforcing 
Steel 


4 
5 
6 

7 


3 

m 

4 
4H 


I H*4r-7H' o.c. 

V0-6H* o.c. 
^8*0-5^' o.c. 




8 
9 

10 
11 


5 

5H 

6 

6H 


H'«-7H'o.c. 

H '0-6K> * o.c. 

y 2 '4>-y o.c. 




12 
13 
14 
15 


7 

8 
9 


^'0-7* o.c. 
^'^-6H'o.c. 
^*0-6' o.c. 
5^*0-5M ' o.c. 




16 
17 
18 
19 
20 


9J4 
10 

1QH 

11H 

12 


J^V-7* o.c. 
^'«-6M'o.c. 
H"4r-6' o.c. 

3 i'V5^'o.c. 





In stair flights which incorporate small platforms, care should be taken 
that the reinforcing steel is properly placed at the junction with the steps, 
particularly where the platform is at the top of the flight. Fig. 19 clearly 
shows this detail. 

r landing 

as ir 

; J 



l"nosfng 




("clear 



Total depth of slab 



DETAIL OF STAIR AND LANDING 
Fig. 19 



Methods of Anchoring Reinforcement 
in Discontinuous Slabs 

There are several ways of anchoring bars at the edges of discontinuous 
slabs or beams. One is to extend the bar into the concrete of the support 
for such a distance that the stresses in the bar are gradually transferred 
by bond to the concrete of the support. 

In cases where the support is not large enough to permit a bar to develop 
its bond, a hook should be provided at the end. Fig. 20 illustrates the right 



84 



Simplified Design of 



— Full s treng th o f 
bar at this section 



,, _■ J. 



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Construction 
joint 



CORRECT METHOD 



1 ' - - ' 



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Crack may open 



* i 



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Construction 
joint 



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X 







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or be continuous 



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f may open 



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CORRECT METHOD 



WRONG METHOD 



Fig. 20 



and wrong method of providing such anchorage. The effectiveness of a 
hook depends partly upon the strength of the concrete. 

The ideal anchorage is obtained by a combination of straight embed- 
ment equal to 15 diameters of the bar and a semi-circular hook, the inside 
diameter of which is at least 4 times the diameter of the bar. With such a 
detail, the elastic limit of the steel can be reached without causing excessive 
stresses in the concrete. 



Construction Joints 

Construction joints will be necessary when the placing of concrete can- 
not be done continuously. They should be near the center of spans of 
beams and slabs, where the shear is a minimum. Bulkheads at right angles 
to the reinforcement prevent the last concrete placed from "feathering" 
out and provide a good surface for the further deposit of concrete. In 
ribbed slabs with metal pans, the bulkheads should be perpendicular to 
the joists. If they are set in the opposite direction, cracks sometimes 
develop at the joints. 



Concrete Floor Systems 



85 



TABLE No. 24 — RECOMMENDED LIVE LOADS 



Apartments .... 
Armories .... 

Auditoriums — fixed seats . 
Auditoriums — movable seats 
Balconies and galleries — fixed seats 
Balconies and galleries — movable seats 

Dance halls .... 
Dwellings .... 

Exterior balconies 
Fire escapes .... 
Garages — all types of vehicles 

Garages — passenger cars only 
Gymnasiums .... 
Hospitals — wards and rooms 
Hospitals — corridors and public rooms 
Hotels — guest rooms and corridors 



Hotels — public rooms and public corridors 
Libraries — reading rooms . 
Libraries — corridors .... 
Loft buildings .... 

Manufacturing — light 

Manufacturing — heavy 

Offices ...... 

Printing plants — composing and linotype rooms 
Public rooms ...... 

Reviewing stands, bleachers, grandstands, etc. 



Roof loads — rise less than 1 to 4 

Schools — class rooms 

Schools — corridors 

Sidewalks — 800 lb. concentrated or 

Stairways 

Storage — light . 
Storage — heavy, not less than 
Stores — retail (light merchandise) 
Stores — wholesale (light merchandise) 
Theatres — stage floor 






Lb. Per 

Sq. Ft. 

40 
150 

50 
100 

50 
100 

120 
40 
100 
100 
100 

80 
100 

40 
100 

40 

100 

50 

100 

100 

75 



125 
50 
100 
100 
100 

30 

50 

100 

250 

100 

100 
250 
75 
100 
100 



Mi 



Simplified Design of 



FABLE No. 25— WEIGHTS OF BUILDING MATEKI \LS 



Asphalt (2" thick) 

Brickwork, common (per inch of thickness) 

Cement floor finish (1" thick) 

Cinders (1" thick) . 

Composition roof . 



Concrete, cinder (1" thick) 
Concrete, stone or gravel (1" th 
Continuous steel sash, glazed 
Creosoted wood blocks (4" thic 
Earth (12" thick) . 

Gypsum (1" thick) 
Hollow clay tile (3* thick) 
Hollow clay tile (4* thick) 
Hollow clay tile (6* thick) 

Hollow clay tile (8" thick) 
Hollow clay tile (10" thick) 

Hollow chi\ tile (12" thick) 
Lightweight Concrete Tile- 
Lightweight ( kmcrete Tile- 
Lightweight Concrete Tile- 



k) 



5* 



ick) 



& 12" 

S"\ 12" 



me one masonry (12* thick) 

Marble mi >nry i 12" thick) 
Partitions, 2 i i Btud, plasU n d both 

Partitions, solid plaster _ thick; 

Plaster, fin tile or concrete 
Plaster, oo n I lath 

lished plate ^1 

Prei ! ceilingt 

Quarry til* 1" thick) 

• 1 <r thick) 
SI ithinfi V P r thick) 

Sidewalk lights 

Skylights, metal, uith v\ire glass 
itC r r 

1 sash, gla I with S S .lass 
Terrazzo floor (2" thick) 



sid< 



Lb. Pel 

Sq . Ft . 

25 
10 

12 

10 

7 

12H 
8-10 

20 

100 

3 

15 

Hi 

22 

30 
36 

4(1 
11 

22 

28 



J70 

L6 
20 

5 

10 
3J 

2 

12 

9 

4 

4(1 

6 
6 



... 



Concrete Floor Systems 



87 



PRINTED IN U. S. A. T - 1 9 - 1 M — 6 - 3 /